Math 215 Exam #1 Practice Problem Solutions


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1 Math 5 Exam # Practice Problem Solutions For each of the following statements, say whether it is true or false If the statement is true, prove it If false, give a counterexample (a) If A is a matrix such that A(Ax) = for all x R, then A is the zero matrix Answer: False If A(Ax) = for all x, then the column space of A and the nullspace of A must be the same space In particular, consider the matrix Then, for any x = and x A = R x, we have that Ax = A(Ax) = x = x x x = Hence, A(Ax) = for all x, but A, so A gives a counterexample to the statement (b) A system of equations in 4 unknowns can never have a unique solution Answer: True We can realize such an system of equations as a single matrix equation Ax = b, where A is a 4 matrix Hence, rank(a), so the dimension of the nullspace of A is at least : dim nul(a) = 4 rank(a) 4 = Hence, there must be at least one free variable in the system, meaning that, if the system is solvable at all, it must have an infinite number of solutions (c) If V is a vector space and S is a finite set of vectors in V, then some subset of S forms a basis for V Answer: { } False Let V = R, which is clearly a vector space, and let S be the singleton set The single element of S does not span R, so no subset of S can be a basis for R Hence, this provides a counterexample to the statement (d) Suppose A is an m n matrix such that Ax = b can be solved for any choice of b R m Then the columns of A form a basis for R m Answer: False Consider the matrix A = Then A is already in reduced echelon form and clearly has pivots, so rank(a) = This implies that dim col(a) =, so the column space of A consists of all of R Thus, the equation Ax = b can be solved for any b R (since any b is in col(a)) However, the columns of A are clearly not linearly independent (no set containing the zero vector can be linearly independent), so they cannot form a basis for R A related but true statement would be the following: Suppose A is an m n matrix such that Ax = b can be solved for any choice of b R m Then some subset of the columns of A forms a basis for R m
2 (e) Given equations in 4 unknowns, each describes a hyperplane in R 4 If the system of those equations is consistent, then the intersection of the hyperplanes contains a line Answer: True This is really just a restatement of (b) Translating the system of equations into a matrix equation Ax = b, the nullspace of A must be at least onedimensional, so the solutionspace must be at least onedimensional Since the solution space of the matrix equation corresponds to the intersection of the hyperplanes, that intersection must be at least onedimensional, meaning it must contain a line (f) If A is a symmetric matrix (ie A = A T ), then A is invertible Answer: False Consider the symmetric matrix A = Then A only has rank, meaning that A cannot be invertible, so this gives a counterexample to the statement (g) If m < n and A is an m n matrix such that Ax = b has a solution for all b R m, then there exists z R m such that Ax = z has infinitely many solutions Answer: True The fact that Ax = b has a solution for all b R m means that the column space of A is equal to all of R m Hence, Since rank(a) = dim col(a) = m dim nul(a) = n rank(a) = n m and since m < n, we have that the nullspace of A has some positive dimension Since the nullspace of A consists precisely of those x R n such that Ax =, this equation has infinitely many solutions Thus, letting z =, we see that the statement is true (h) The set of polynomials of degree 5 forms a vector space Answer: True You should check that the set of polynomials of degree 5 satisfies all the rules for being a vector space The important facts are this space is closed under addition and scalar multiplication For each of the following, determine whether the given subset is a subspace of the given vector space Explain your answer (a) Vector Space: R 4 Subset: The vectors of the form Answer: Yes, this is a subspace If we take two vectors in the subset, say a b d a b and a b, then their sum a b d + a b d = is also in the subset, so this set is closed under addition a + a b + b d + d d d
3 Moreover, if c R, then c a b d = is in the set, so this set is closed under scalar multiplication Thus, the set is closed under both addition and scalar multiplication, and so is a subspace (b) Vector Space: R Subset: The solutions to the equation x 5y = Answer: No, this is not a subspace To see why, I ll show that it is not closed under addition The vectors and are both in the set, since the pairs (/, ) and (, /5) both solve the equation x 5y =, but + = 5 5 is not in the set, since ca cb cd 5 (/) 5( /5) = + = Therefore, the set is not closed under addition, and so is not a subspace (c) Vector Space: R n Subset: All x R n such that Ax = x where A is a given n n matrix Answer: Yes, this is a subspace To prove it, suppose x and x are in this set, meaning that Ax = x and Ax = x (such vectors are called eigenvectors of A; we ll learn more about them later) Then A(x + x ) = Ax + Ax = x + x = (x + x ), meaning that x + x is in this set as well Moreover, for any c R, A(cx ) = c(ax ) = c(x ) = (cx ), so cx is in the set as well Therefore, this set is closed under addition and scalar multiplication, so it is indeed a subspace (d) Vector Space: R Subset: The intersection of P and P, where P and P are planes through the origin Answer: Yes, this is a subspace The proof is essentially the same as you gave for Problem (c) from HW 4 (e) Vector Space: All polynomials Subset: The quadratic (ie degree ) polynomials Answer: No, this is not a subspace To see that it is not closed under addition, notice that if f(t) = t and g(t) = t, then f and g are both in the set of quadratic polynomials, but, since the sum f + g is not a quadratic polynomial (f + g)(t) = f(t) + g(t) = t + ( t ) =,
4 (f) Vector Space: All realvalued functions Subset: Functions of the form f(t) = a cos t + b sin t + c for a, b, c R Answer: Yes, this is a subspace If a, a, b, b, c, c R and I define and f(t) = a cos t + b sin t + c g(t) = a cos t + b sin t + c, then f and g are in the given subset The sum has the form f(t)+g(t) = (a cos t+b sin t+c )+(a cos t+b sin t+c ) = (a +a ) cos t+(b +b ) sin t+(c +c ), so f + g is also in the subset, which is, therefore, closed under addition Also, if r R, then rf(t) = r(a cos t + b sin t + c ) = (ra ) cos t + (rb ) sin t + (rc ), so rf is in the subset, which is, therefore, closed under scalar multiplication Hence, we can conclude that this subset is actually a subspace Consider the matrix A = a a (a) Under what conditions on a is A invertible? Answer: The matrix A is invertible if and only if it has rank To see what the rank is, we do elimination The first step is to subtract a times row from row, yielding a a Then this has a second pivot if and only if a, meaning that a, or a ± Thus, A is invertible so long as a is neither nor (b) Choose a nonzero value of a that makes A invertible and determine A Answer: Choose a = Recall that we can find the inverse of A by converting the left side of the following augmented matrix to the identity: Subtract twice row from row : Scale the second row by Therefore, and also subtract twice the result from row : A = 4
5 (c) For each value of a that makes A noninvertible, determine the dimension of the nullspace of A Answer: When a =, the matrix A =, which, after subtracting row from row, reduces to Hence, A has rank, so the nullspace has dimension 4 Consider the system of equations dim nul(a) = rank(a) = = x + x + x x 4 = b x + x + x 5x 4 = b x + 4x + x 8x 4 = b (a) Find all solutions when the above system is homogeneous (ie b = b = b = ) Find a basis for the space of solutions to the homogeneous system Answer: Convert the system into the augmented matrix Now do elimination to get the reduced echelon form First, subtract row from row and subtract twice row from row : Now, subtract row from both row and row : Then this system is consistent provided that x = x + x 4 x = x 4 Hence, the solutions to the homogeneous equation are those vectors of the form x + x 4 for x, x 4 R Then a basis for the space of solutions to the homogeneous system (ie nullspace of the corresponding matrix) is, 5
6 (b) Let S be the set of vectors b = b b such that the system can be solved What is the dimension b of S? Answer: Letting A be the matrix of the system, we know that the set of vectors b for which the system can be solved is the column space of A Since A is 4, we know that rank(a) + dim nul(a) = 4 Since, from part (a), we know that the dimension of the nullspace is, this implies that the column space of A is twodimensional (c) It s easy to check that the vector v = is a solution to the system that arises when b =, b = 5, and b = 8 Find all the solutions to this system Answer: All solutions x to the system Ax = b take the form x = x p +x h, where x p is a particular solution and x h is the homogeneous solution to the corresponding homogeneous problem Thus, we can let x p = v, which we re told solves the system and we see that, using part (a), the general solution is where x, x 4 R + x + x 4, 6
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